PII. S0161171201005348 http://ijmms.hindawi.com © Hindawi Publishing Corp.
A FUNCTIONAL EQUATION CHARACTERIZING
CUBIC POLYNOMIALS AND ITS STABILITY
SOON-MO JUNG and PRASANNA K. SAHOO
(Received 30 May 2000)
Abstract.We study the generalized Hyers-Ulam stability of the functional equation
f [x1, x2, x3]=h(x1+x2+x3).
2000 Mathematics Subject Classification. 39B22, 39B82.
1. Introduction. Given an operatorT and a solution class{u}with the property that T (u)=0, when does T (v) ≤εfor an ε >0 imply that u−v ≤δ(ε) for some uand for some δ >0? This problem is called the stability of the functional transformation. A great deal of work has been done in connection with the ordinary and partial differential equations. Iffis a function from a normed vector space into a Banach space, andf (x+y)−f (x)−f (y) ≤ε, Hyers [3] proved that there exists an additive mapAsuch thatf (x)−A(x) ≤ε. Iff (x)is a real continuous function ofxoverR, and|f (x+y)−f (x)−f (y)| ≤ε, it was shown by Hyers and Ulam [4] that there exists a constantksuch that|f (x)−kx| ≤2ε. Taking these results into account, we say that the additive Cauchy equationf (x+y)=f (x)+f (y)is stable in the sense of Hyers and Ulam.
In this paper, we study a generalized Hyers-Ulam stability of a mean value type functional equation.
LetRbe the set of real numbers. For distinct pointsx1, x2, . . . , xninR, the divided difference off:R→Ris recursively defined as
fx1=fx1,
fx1, x2, . . . , xn=f
x1, x2, . . . , xn−1−fx2, x3, . . . , xn
x1−xn .
(1.1)
Bailey [2], generalizing a result of Aczel [1], proved the following result: if f is a differentiable function satisfying the functional equation
fx1, x2, x3=hx1+x2+x3, ∀x1, x2, x3∈R (1.2)
with x1≠x2, x2≠x3, x3≠x1, thenf is a polynomial of degree at most three. In Bailey’s proof, the differentiability assumption plays a central role. Kannappan and Sahoo [5] have determined the general solution off [x1, x2, . . . , xn]=h(x1+x2+···+
2. Solution of the functional equation (1.2). Now we give the solution of the func-tion equafunc-tion (1.2) using an elementary technique.
Theorem2.1. Letf satisfy the functional equation (1.2) for allx1, x2, x3∈Rwith
x1≠x2,x2≠x3, andx3≠x1. Thenf is a polynomial of degree at most three andh is linear.
Proof. Iff (x)is a solution of (1.2) so isf (x)+a0+a1x, where a0and a1are arbitrary constants. This can be verified by direct substitution into the expansion of the functional equation (1.2), that is,
x2−x3fx1+x3−x1fx2+x1−x2fx3
=x1−x3x1−x2x2−x3hx1+x2+x3. (2.1)
Lettingf (xi)+a0+a1xifori=1,2,3 forf (xi)in the expansion (2.1), we get
x2−x3fx1+a0+a1x1+x3−x1fx2+a0+a1x2
+x1−x2fx3+a0+a1x3
=x1−x3x1−x2x2−x3hx1+x2+x3.
(2.2)
Each term involving ana0or ana1has an opposite-sign term and therefore cancels by simple algebraic manipulation. Thus we have again the expanded form (2.1) of (1.2). Letg(x)=f (x)+a0+a1x. Thenx=0 inserted intog(x)yields
f (0)=g(0)−a0. (2.3)
We are free to picka0=g(0)so thatg(x)yieldsf (0)=0. In other words, by a suitable choice fora0, without loss of generality, we may assume that
f (0)=0. (2.4)
Now by settingx=αin the definition ofg(x)we get
f (α)=g(α)−a0−a1α. (2.5)
Lettinga0+a1α=g(α)we getf (α)=0 and we may assume, without loss of gener-ality, that
f (α)=0 (2.6)
for someα≠0 inR. Note that there are many choices for such anα. First substitute(x,0, α)for(x1, x2, x3)in (2.1) to get
f (x)= −x(α−x)h(x+α) (2.7)
(after using (2.4) and (2.6)) forx≠0, α.
Next, we substitute(x,0, y)for(x1, x2, x3)in (2.1) to get
f (x) x(x−y)−
f (y)
Define
g(x)=f (x)x forx∈R\{0}. (2.9)
Then (2.8) reduces to
g(x)−g(y)=(x−y)h(x+y) ∀x, y∈R\{0}withx≠y. (2.10)
Note that (2.10) is valid even forx=y. Now we consider the equation
g(x)−g(y)=(x−y)h(x+y) ∀x, y∈R\{0}. (2.11)
Puty= −xin (2.10) to get
g(x)−g(−x)=2xh(0) ∀x≠0. (2.12)
Next, replaceyby−yin (2.10) to get
g(x)−g(−y)=(x+y)h(x−y) forx, y∈R\{0}withx+y≠0. (2.13)
Again (2.13) holds ifx+y=0. Thus we conclude that (2.13) holds forx, y∈R\{0}. Subtract (2.10) from (2.13) and use (2.12) to get
(x+y)h(x−y)−h(0)=(x−y)h(x+y)−h(0) ∀x, y∈R\{0}. (2.14)
Fix a nonzerouinR. Choose av∈Rsuch that(u+v)/2≠0 and(u−v)/2≠0. There are plenty of choices for suchv. Let
x=u+2v, y=u−2v, (2.15)
so that
u=x+y, v=x−y. (2.16)
Letting (2.16) into (2.14), we get
uh(v)−h(0)=vh(u)−h(0) ∀v≠u,−u. (2.17)
(Here note thatvcan be zero sincex=yis allowed.) Hence for fixedu=u1, we get
h(v)=a1v+b1 forv∈R\u1,−u1. (2.18)
Againu=u2, we get
h(v)=a2v+b2 ∀v∈R\u2,−u2. (2.19)
Since the sets{u1,−u1}and{u2,−u2}are disjoint, we get
h(v)=av+b ∀v∈R. (2.20)
Now using (2.20) in (2.7), we have
wherec= −aα2−bα. Removing the assumption thatf (0)=0, we get
f (x)=ax3+bx2+cx+d ∀x≠0, α. (2.22)
By (2.4), (2.6), and (2.22), we conclude thatf is a polynomial of degree at most three for allx∈R. This proof is now complete.
For a more general result, the interested reader should refer to Kannappan and Sahoo [5].
3. Stability of the functional equation (1.2). LetGbe an additive subgroup ofC
and letϕ:G3→[0,∞)be a control function. In the following theorem, the stability of (1.2) for cubic polynomials will be investigated in a modified form (3.1).
Theorem 3.1. Let α∈ G\ {0} and β∈G\ {−α,0, α} be fixed. If the functions
f , h:G→Csatisfy the inequality
(y−z)f (x)+(z−x)f (y)+(x−y)f (z)
−(x−z)(x−y)(y−z)h(x+y+z)≤ϕ(x, y, z), ∀x, y, z∈G, (3.1)
then there exist constantsa, b, c, dsuch that
f (x)−ax3−bx2−cx−d≤ x2−α2
2|β|β2−α2ϕ(x, β,−β)
+ x2−β2
2|α|β2−α2ϕ(x, α,−α) ∀x∈G,
(3.2)
|h(x)−ax−b| ≤ x
2−β2+β2−α2
2|α|β2−α2x2−α2ϕ(x, α,−α)
+ 1
2|β|β2−α2ϕ(x, β,−β) ∀x∈G\{−α, α}.
(3.3)
Moreover, the constantsa, b, c, dare explicitly given by
a=f (β)2β−f (−β) β2−α2 −
f (α)−f (−α) 2αβ2−α2 ,
b=f (β)2 +f (−β) β2−α2 −
f (α)+f (−α) 2β2−α2 ,
c=f (α)−f (−α) 2αβ2−α2 β
2−f (β)−f (−β) 2ββ2−α2 α
2,
d=f (α)+f (−α) 2β2−α2 β
2−f (β)+f (−β) 2β2−α2 α
2.
(3.4)
Proof. If we define a functiong:G→Cby
theng(α)=g(−α)=0 andgsatisfies the inequality (y−z)g(x)+(z−x)g(y)+(x−y)g(z)
−(x−z)(x−y)(y−z)h(x+y+z)≤ϕ(x, y, z) ∀x, y, z∈G. (3.6)
If we substitute(x, α,−α)for(x, y, z)in (3.6), then we have
g(x)−
x2−α2h(x)≤ 1
2|α|ϕ(x, α,−α) ∀xinG. (3.7)
Replacezby−yin (3.6) to get
2yg(x)−(x+y)g(y)+(x−y)g(−y)−2y
x2−y2h(x)≤ϕ(x, y,−y) (3.8)
for everyx, y∈G.
By making use of (3.7) and (3.8), we obtain
2yg(x)−(x+y)g(y)+(x−y)g(−y)−2y
x2−y2 x2−α2 g(x)
≤2yg(x)−(x+y)g(y)+(x−y)g(−y)−2yx2−y2h(x)
+2yx2−y2h(x)−2y
x2−y2 x2−α2 g(x)
≤ϕ(x, y,−y)+|y|x 2−y2
|α|x2−α2ϕ(x, α,−α)
(3.9)
or equivalently
2yy 2−α2
x2−α2g(x)−(x+y)g(y)+(x−y)g(−y)
≤ϕ(x, y,−y)+||y|x2−y2
α|x2−α2ϕ(x, α,−α), ∀x∈G\{−α, α},∀y∈G. (3.10)
Multiply both sides by
x2−α2
2|y|y2−α2 (3.11)
to get g(x)−
x2−α2(x+y) 2yy2−α2 g(y)+
x2−α2(x−y) 2yy2−α2 g(−y)
≤ x2−α2
2|y|y2−α2ϕ(x, y,−y)
+ x2−y2
2|α|y2−α2ϕ(x, α,−α) ∀x∈G, y∈G\{−α,0, α}.
(3.12)
(We note that the inequality holds true also forx∈ {−α, α}.)
By using (3.2), (3.5), and (3.7), we may obtain x2−α2h(x)−ax3−bx2+α2ax+α2b
≤x2−α2h(x)−g(x)
+g(x)−f (x)+f (α)−2αf (−α)x+f (α)+2f (−α)
+f (x)−ax3−bx2−cx−d
≤x2−β2+β2−α2
2|α|β2−α2 ϕ(x, α,−α)+
x2−α2
2|β|β2−α2ϕ(x, β,−β) ∀x∈G, (3.13)
from which we can deduce inequality (3.3).
Corollary 3.2. Assume that the control function ϕ :G3 →[0,∞) satisfies the asymptotic condition
lim |x|→∞|x|
2ϕ(x, γ,−γ)=0 for each fixedγ∈G. (3.14)
If the functionsf , h:G→Csatisfy inequality (3.1) for anyx, y, z∈G, then there exist uniquely determined constantsa, b, c, dsuch that inequalities (3.2) and (3.3) are valid for allx∈Gand for allx∈G\{−α, α}, respectively.
Corollary3.3. Suppose that the control functionϕ:G3→[0,∞)is given by
ϕ(x, y, z)=ε|x−y||y−z||z−x| for some givenε >0. (3.15)
If the functionsf , h:G→Csatisfy inequality (3.1) for anyx, y, z∈G, then there exist constantsa, b, c, dsuch that
f (x)−ax3−bx2−cx−d≤ 2ε β2−α2x
2−α2x2−β2,
h(x)−ax−b≤ε+ 2ε
β2−α2x
2−β2 ∀xofG.
(3.16)
We remark here that the last inequality is also valid forx= −αorx=α.
Given a control functionψ:G3→[0,∞), we can also prove the Hyers-Ulam-Rassias stability of the functional equation (1.2) in the original setting:
Theorem 3.4. Let α∈G\ {0} and β∈G\ {−α,0, α} be given. If the functions
f , h:G→Csatisfy the inequality
f [x, y, z]−h(x+y+z)≤ψ(x, y, z) ∀x, y, z∈Gwithx≠y, y≠z, z≠x, (3.17)
then there exist constantsa, b, c, dsuch that
f (x)−ax3−bx2−cx−d≤x2−α2x2−β2 β2−α2
ψ(x, α,−α)+ψ(x, β,−β), (3.18)
|h(x)−ax−b| ≤ψ(x, α,−α)+βx22−−αβ22
ψ(x, α,−α)+ψ(x, β,−β), (3.19)
Proof. If we multiply both sides of (3.17) by|x−y||y−z||z−x|, thenfsatisfies
inequality (3.1) with
ϕ(x, y, z)= |x−y||y−z||z−x|ψ(x, y, z) ∀x, y, z∈G. (3.20)
(We note that (3.1) is also true forx, y, z∈Gwithx=y,y=z, orz=xfor our case with (3.20).)
According to Theorem 3.1, there exist constants a, b, c, d such that inequalities (3.18) and (3.19) are valid for all x ∈G and for all x ∈G\ {−α, α}, respectively. The only reason for excepting−αand αfrom the domain of validity of inequality (3.3) is that the denominator of the first term on the right-hand side contains a fac-tor|x2−α2|. However, inequality (3.19) contains no denominator which vanishes at x=αorx= −α. Therefore, we can include−αandαin the domain of validity of inequality (3.19), which completes the proof.
References
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Soon-Mo Jung: Mathematics Section, College of Science & Technology, Hong–Ik University,339–701Chochiwon, Korea
E-mail address:[email protected]
Prasanna K. Sahoo: Department of Mathematics, University of Louisville, Louisville, KY40292, USA